# Finite non-abelian group of the order pq, pq primes

#### Sportin' Life

Howdy, all. I am not sure if this is the right forum for this question. It isn't exactly an homework question, but it does stem very closely from a homework assignment in a first year graduate course in Abstract Algebra. The assignment has come and gone with limited success on my part, but some questions remain.

The problem was the well known one about a finite group G, |G| = pq where p<q primes. The object was to use Sylow Theorems to show that there is a unique (up to isomorphism) non-abelian group G if, and only if, p|q-1. Of course it is simple to show the case where p does not divide q-1, and my problem came from proving the other case. My question about that case is: Can one prove that there is such a unique non-abelian group without using any theory of automorphisms? Specifically is there a clever group action that can deliver the desired result?

The reason I ask is that our course has not mentioned automorphisms one bit. The text we are using is Ash Abstract Algebra: The Basic Graduate Year, though most of our problems are pulled from Herstein's Topics in Algebra. I have managed to read through the Herstein and pull out the tools I need for the problem, but could I have done the problem with the theory that has so far been provided? So far in a nutshell we have had the isomorphism theorems, the definition of group action, the orbit stabilizer theorem, and the class equation.

Thank you for your time. Cheers.

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#### fresh_42

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2018 Award
The Abelian case is $G \cong \mathbb{Z}_p \times \mathbb{Z}_q$, which means for a non Abelian group G we will have a semidirect product, which means one factor operates via an automorphism on the other one. Thus we will automatically have to deal with an automorphism. Now if you do not want to name it, you will have to use the mapping.

The question comes down to: Which one of the two factors is the normal subgroup and how does a non trivial conjugation of the non normal subgroup on the normal one looks like?
This gives you at least the action you are looking for.

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